TITLE: Disproving an "almost true" trigonometric identity
QUESTION [27 upvotes]: The plausible looking "identity"
$$\sin(\frac{\pi}{51})+\cos(\frac{\pi}{74})=\frac{3}{2\sqrt 2}$$
is not true, but it is close indeed:
$$LHS=1.0606\color{blue}{598...}$$
$$RHS=1.0606\color{red}{601...}$$
In fact the difference, on the order of $10^{-7}$, is small enough to fool some cheap calculators.
Is there any way to disprove this identity without relying on a (good) calculator?

REPLY [15 votes]: First note that $[\mathbb{Q}(\sin\tfrac{\pi}{51}):\mathbb{Q}]=32$ so $[\mathbb{Q}(\sin\tfrac{\pi}{51}, \sqrt{2}):\mathbb{Q}]\in\{32,64\}.$ Now $[\mathbb{Q}(\cos\tfrac{\pi}{74}):\mathbb{Q}]=36$ and since $36\not\mid 64$ we conclude that $\cos\tfrac{\pi}{74}\not\in\mathbb{Q}(\sin\tfrac{\pi}{51}, \sqrt{2})$ and in particular $\cos\tfrac{\pi}{74}\neq \frac{3}{2\sqrt{2}}-\sin\tfrac{\pi}{51}$.